7 Java与UML协同应用开发.ppt

杨海秀 yanghaixiu09@ 面向对象与UML 第 七章 Java与UML协同应用开发 根据最短路径计算基因集合功能相似性 生物学意义：节点间路径越短，说明参与同一个功能的可能性越大。 需求分析： 开发一个程序，能够通过计算两组数据间每对节点的最短路径对基因集合功能相似性进行打分，需要具有以下功能： 输入两组基因集合，计算两组基因集合的功能相似性 输入两个文件，计算两个文件中每两组基因集合的相似性 根据如上问题描述进行用例建模。 问题描述：构建无权无向图，投入两组基因，通过计算两组基因中每对基因的最短路径对两组基因进行打分(算法已知)，则分值越高，功能越相似。 we demonstrated how to compute the CCRS of two gene sets, G1 and G2. The intersection of G1 and G2 is denoted by G. There are n paths among the m nodes of the intersection G, based on the PPI network, for two genes p and q that are taken from G1 and G2, respectively. Where rpq represents the ‘function distance’ (the length of the shortest path) between gene p and gene q, and Rpq is the rank score between them, Rpq= 1/rpq. N is the number of the existing ‘function distance’ values (including the ‘function distance’ of the overlapping nodes (m) from the node to itself). When there is no path from gene p to gene q, we can define rpq=∞ and in this case, the rank score (Rpq) is zero. In our study, we set α=β=γ=δ=1. In the definition of the CCRS method, we learn that Σp,q∈G α·Rpq = Σp,q∈G (1/rpq) = m + n. 算法描述： 两组基因5700,5701,5702,5704,5705,5706,5707,5716,5717,5718,5719,10213,5708,5709,5710,5711,9861,5713,5714,5715 57122,55746,23279,79023,348995,79902,4928,6396,81929 计算两组中每对基因的最短路径，取最短路径=4的基因对。 Thank you!